Question

A wooden block of mass $0.5kg$ and density $800kg{m}^{-3}$ is fastened to the free end of vertical spring of spring constant $500N{m}^{-1}$ fixed at the bottom. If the entire system is completely immersed in water, find (a) the elongation (or compression) of the spring in equilibrium and (b) the time-period of vertical oscillations of the block when it is slightly depressed and released.

Answer 1

Given

mass of wooden block $m=0.5kg$

Density of block $\rho =800kg/m^2$

Spring constant $k=50N/m$

Distance of water $={\rho }_w=1000kg/m^3$

$\left(a\right)$

$mg=kx+v{\rho }_{w}g$

$(0.5\times (10+50x\left)\right)=\frac{0.5}{800}\times {10}^3\times 10$

$50x=0.5\times 10\times (\frac{10}{5}-1)\Rightarrow 50x=5/4$

$x=2.5cm$

$\left(b\right)$

As the system insiden water. The Unbalance force will be the deiving force what is $kx$ for $SHM$ hence

There will be no chage in between force

$ma=kx$

$a=\frac{kx}{m}$

$w^{2}x=\frac{kx}{m}\Rightarrow {\left(\frac{2\pi }{T}\right)}^2=\frac{k}{m}$

$T=2\pi \sqrt{\frac{m}{k}}$

$=2x\sqrt{\frac{0.5}{50}}=\frac{\pi }{5}s$